I'm trying to generate phase data from magnitude data in a frequency function, assuming the system is minimum phase. Using Hilbert Transform.

For instance, having this simple system:

$G(s) = s$

$G(j\omega) = j\omega$

Magnitude is: $|G(j \omega)| = |j \omega| = \omega$

And phase is: $\arg[G(j\omega)] = \arg[j\omega] = \frac\pi2$

In spite on knowing the system's phase response. I want to calculate the phase using Hilbert Transform.

In this wikipedia article, we find the magnitude and phase relationship of a minimum phase system, using Hilbert transform.

$\arg \left[ G(j \omega) \right] = -\mathcal{H} \lbrace \ln \left( |G(j \omega)| \right) \rbrace $

So for our example:

$ \arg \left[ G(j \omega) \right] = -\mathcal{H} \lbrace \ln (\omega) \rbrace = \frac\pi2 $

ln is base e logarithm.

Being the Hilbert Transform defined as:

$\mathcal{H} \lbrace G(\omega) \rbrace \ \stackrel{\mathrm{def}}{=}\ \widehat{G}(\omega) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{G(\tau)}{\omega-\tau}\, d\tau $

So, for the system:

$ -\mathcal{H} \lbrace \ln (\omega) \rbrace = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\ln(\tau)}{\omega-\tau}\, d\tau = \frac{\pi}2 $

I need help for solve the improper integral. The anti-derivative of the expression inside the integral seems to be quite complicated. How do I use the Cauchy principal value?

How is the solution developed? How do I demostrate this equality? Thanks


1 Answer 1


The problem with your example is that there's a zero right on the imaginary axis, making the system not strictly minimum-phase, and inversion with a causal and stable filter is not possible. So let's slightly modify the frequency response by introducing a small constant $\epsilon >0$:


This makes sure that the zero of $G(s)$ is strictly in the left half-plane, making the system strictly minimum-phase. Magnitude and phase are given by

\begin{eqnarray*} |G(j\omega)|&=&\sqrt{\epsilon^2+\omega^2}\\ \arg\{G(j\omega)\}&=&\arctan\left(\frac{\omega}{\epsilon}\right) \end{eqnarray*}

and we're trying to show that



\begin{eqnarray*} \ln\left(\sqrt{\epsilon^2+\omega^2}\right)&=&\frac12\ln\left(\epsilon^2+\omega^2\right)\\ &=&\frac12\ln\left(\epsilon^2\right)+\frac12\ln\left(1+\left(\frac{\omega}{\epsilon}\right)^2\right) \end{eqnarray*}

we see that


because the Hilbert transform of a constant is zero.

In this answer I showed that


Using $(3)$, a simple substitution in $(4)$ shows that $(2)$ is indeed true.

Note that magnitude and phase of the originally proposed system - a differentiator - are given by

\begin{eqnarray*} |G(j\omega)|&=&|\omega|\\ \arg\{G(j\omega)\}&=&\frac{\pi}{2}\textrm{sign}(\omega) \end{eqnarray*}

From the correspondence $(2)$ we see that for $\epsilon\to 0$ the left-hand side becomes the Hilbert transform of $-\ln(|\omega|)$ and the right-hand side converges to $\frac{\pi}{2}\textrm{sign}(\omega)$.

It turns out that it can even be shown directly that $-\ln|\omega|$ and $\frac{\pi}{2}\textrm{sign}(\omega)$ are a Hilbert transform pair. First note that




are equivalent statements because $\mathcal{H}^{-1}\{\cdot\}=-\mathcal{H}\{\cdot\}$.

In this answer I proved Eq. $(6)$. A short and simple version of the derivation is this:

\begin{eqnarray*} \mathcal{H}\left\{\frac{\pi}{2}\textrm{sign}(\omega)\right\}&=&\frac{\pi}{2}\frac{1}{\pi\omega}\star\textrm{sign}(\omega)\\&=&\frac12\left[\ln|\omega|\right]'\star\textrm{sign}(\omega)\\&=&\frac12\ln|\omega|\star\left[\textrm{sign}(\omega)\right]'\\&=&\frac12\ln|\omega|\star 2\delta(\omega)\\&=&\ln|\omega| \end{eqnarray*}

where $\star$ denotes convolution, and $'$ denotes the derivative.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.